What is the integral of (4e^4)^x ?

The integral of (4e^4)^x is 1/(ln(2)+2)*2^{2x-1}e^{4x}+C

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integral of (4e^4)^x

\int (4 e^{4})^{x} d x

\frac{1}{ln ( 2 ) + 2} \cdot 2^{2 x - 1} e^{4 x} + C

Solution steps

\int (4 e^{4})^{x} d x

\int a^{x} d x = \frac{a ^{x}}{ln a}

= \frac{( 4 e ^{4} ) ^{x}}{ln ( 4 e ^{4} )}

Simplify

\frac{( 4 e ^{4} ) ^{x}}{ln ( 4 e ^{4} )} : \frac{1}{ln ( 2 ) + 2} \cdot 2^{2 x - 1} e^{4 x}

= \frac{1}{ln ( 2 ) + 2} \cdot 2^{2 x - 1} e^{4 x}

Add a constant to the solution

= \frac{1}{ln ( 2 ) + 2} \cdot 2^{2 x - 1} e^{4 x} + C

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What is the integral of (4e^4)^x ?
The integral of (4e^4)^x is 1/(ln(2)+2)*2^{2x-1}e^{4x}+C